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Find a function less than a fraction of itself

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Problem Statement
Give an example of a function ##u## such that ##u:I\to I=(−1,1)## such that ##|u(x)−u(y)|\leq 0.5|x−y|## for all ##x,y\in I##, and ##u(x) \neq x## for all ##x\in I##.
Relevant Equations
Nothing comes to mind
Well doesn't ##u(x) = 0.4 x## work? Seems too easy, but the phrasing at the end "for all ##x\in I##" makes me think since ##0.4x = x## only at ##x=0##, and not all of ##I##, that this is okay. But am I wrong?
 

fresh_42

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##u(x) \neq x## for all ##x\in I## means, that none of the ##x## should have this property. Thus linear functions are ruled out.
 
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##u(x) \neq x## for all ##x\in I## means, that none of the ##x## should have this property. Thus linear functions are ruled out.
But then ##\sin(x)## is also ruled out? And so is ##x^{2}##. Any ideas?
 
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But then ##\sin(x)## is also ruled out? And so is ##x^{2}##. Any ideas?
How about fractional powers? You want a function that is bounded above by y = (1/2)x on [0, 1] and bounded below by the same function on [-1, 0].
 
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How about fractional powers? You want a function that is bounded above by y = (1/2)x on [0, 1] and bounded below by the same function on [-1, 0].
But at ##x=0## all fractional powers are zero. So it won't work, right?
 
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But the function requires ##I\to I##.
I don't think that precludes omitting 0 in how the function is defined.

However, if you don't like that, one can interpret the requirement
and ##u(x) \neq x## for all ##x\in I##.
to mean that u(x) is not identically equal to x. I read this as different from ##\forall x \in (-1, 1), f(x) \ne x##; i.e., for each x in I, ##f(x) \ne x##.
 
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I don't think that precludes omitting 0 in how the function is defined.

However, if you don't like that, one can interpret the requirement to mean that u(x) is not identically equal to x. I read this as different from ##\forall x \in (-1, 1), f(x) \ne x##; i.e., for each x in I, ##f(x) \ne x##.
So is there even such a function that takes ##I\to I##, satisfying the inequality, where it does not cross the line ##y=x## anywhere in ##I##? I also read it that way at first, but it seemed way too easy.
 

PeroK

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So is there even such a function that takes ##I\to I##, satisfying the inequality, where it does not cross the line ##y=x## anywhere in ##I##? I also read it that way at first, but it seemed way too easy.
Can you show that ##u## must be continuous?

Can you draw such a function? Maybe forget formulas to begin with. What would it look like?
 
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PeroK

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PS if ##u## must also be onto, then it is impossible. You might want to try proving this.

Hint: assume first that ##u(0) > 0## and reach a contradiction to the required inequality.
 

fresh_42

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As I read it, the function doesn't need to be surjective, but defined everywhere on ##I=(-1,1)## as the condition involves all points of ##I##. This means you have three tasks:
a) let ##u(x)## be a composite function, e.g. two linear ones with a 'jump' at ##x=0##
b) prove what @PeroK said: there is no such function if it has to be surjective on the closed interval ##[-1,1]##
c) this leads to what is probably meant by the exercise: ##u(x)=x## has to be at the points ##x=\pm 1## for otherwise it is not possible, which leads back to @Mark44 's parabola - just lift it a bit above the origin
 

PeroK

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There is actually a simple solution. A simple family of solutions, in fact.
 
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What about the function ##u(x) = 2##? Then ##|u(x)-u(y)| = |2-2| = 0 \leq 0.5|x-y|##. Additionally, in ##I## we note that ##u(x) \neq x##.
 
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As I read it, the function doesn't need to be surjective, but defined everywhere on ##I=(-1,1)## as the condition involves all points of ##I##. This means you have three tasks:
a) let ##u(x)## be a composite function, e.g. two linear ones with a 'jump' at ##x=0##
b) prove what @PeroK said: there is no such function if it has to be surjective on the closed interval ##[-1,1]##
c) this leads to what is probably meant by the exercise: ##u(x)=x## has to be at the points ##x=\pm 1## for otherwise it is not possible, which leads back to @Mark44 's parabola - just lift it a bit above the origin
A composite linear function with a jump at ##x=0## seems impossible to force ##|u(x)-u(y)| \leq 0.5|x-y|## since the ##\lim_{x \to 0^-}u \neq \lim_{x \to 0^+}u##, yet ##\lim_{x\to 0} 0.5x = 0##. Am I missing something?
 

PeroK

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Given a composite linear function with a jump at ##x=0## seems impossible to force ##|u(x)-u(y)| \leq 0.5|x-y|## since the ##\lim_{x \to 0^-}u \neq \lim_{x \to 0^+}u##, yet ##\lim_{x\to 0} 0.5x = 0##. Am I missing something?
The function must be continuous. The only real constraint is that ##u## must not cross the line ##y=x##.

Why not keep ##u## above that line?

When you tried ##u(x) = 2##, you would have been closer with ##u(x) = 1##.
 
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The function must be continuous. The only real constraint is that ##u## must not cross the line ##y=x##.

Why not keep ##u## above that line?

When you tried ##u(x) = 2##, you would have been closer with ##u(x) = 1##.
Yea, I was about to respond with that but ##1\notin I##.

If we agree the function must be continuous, then are you saying make like a ##V## but not cross ##y=x##? Seems like a good solution!
 

PeroK

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Yea, I was about to respond with that but ##1\notin I##.

If we agree the function must be continuous, then are you saying make like a ##V## but not cross ##y=x##? Seems like a good solution!
You don't need a V.
 

PeroK

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Yea, I was about to respond with that but ##1\notin I##.

If we agree the function must be continuous, then are you saying make like a ##V## but not cross ##y=x##? Seems like a good solution!
Okay. Go for a V. I keep looking at the OP to see whether I'm missing something!
 

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But then ##\sin(x)## is also ruled out? And so is ##x^{2}##. Any ideas?
How about rescaling something that is increasing , with |f(-1)|=|f(1)|? Just rescale by the value at either endpoint.
 
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Wait, I totally spaced the function is ##y=0.5x## and not ##y=x##. In this case ,why not ##u(x) = 0.7##? Doesn't this work?
 

PeroK

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Wait, I totally spaced the function is ##y=0.5x## and not ##y=x##. In this case ,why not ##u(x) = 0.7##? Doesn't this work?
I can't understand the issue here. It's ptobably time to wrap this up and move on. Any linear function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

And, indeed, any differentiable function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

The same for functions that lie below the line ##y = x## and intersect it at ##x = -1##.

Maybe there was something about the working of the problem that confuses everyone. Maybe it was too much focus on formulas rather than a focus on some simple geometric constraints.

The only two constraints were

a) ##|u(x) - u(y)| \le 0.5 |x-y|##

This enforces continuity and means the function can't be too steep.

b) ##u(x) \ne x##

This means that the function must lie above or below the line ##y=x##.

For example, the straight line joining ##(-1, 0)## with ##(1, 1)## would do. It has a constant gradient of ##1/2## and intersects the line ##y=x## outside the open interval ##I##.
 
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I can't understand the issue here. It's ptobably time to wrap this up and move on. Any linear function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

And, indeed, any differentiable function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

The same for functions that lie below the line ##y = x## and intersect it at ##x = -1##.

Maybe there was something about the working of the problem that confuses everyone. Maybe it was too much focus on formulas rather than a focus on some simple geometric constraints.

The only two constraints were

a) ##|u(x) - u(y)| \le 0.5 |x-y|##

This enforces continuity and means the function can't be too steep.

b) ##u(x) \ne x##

This means that the function must lie above or below the line ##y=x##.

For example, the straight line joining ##(-1, 0)## with ##(1, 1)## would do. It has a constant gradient of ##1/2## and intersects the line ##y=x## outside the open interval ##I##.
Yea sorry, I was really spacing out. But thanks! Kinda too late for the HW anyways, buuuuuuut I appreciate you sticking with it!

So if the interval ##I = \mathbb R## and we require the inequality be strictly less than, then would there be any such ##u(x)##?
 
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